2.6 The first and second laws of thermodynamics for continua 

 

 

Consider a sub-region V of a deformed solid with surface A, as shown in the figure.   The solid has mass density $\rho$ (mass per unit deformed volume) Define:

 

· The heat flux vector q flowing through the solid , which is defined so that $dQ=q\cdot ndA$ is the heat flux crossing an internal surface with area dA and normal n in the deformed solid;

 

· The heat supply q , defined so that dQ= qdV is the heat supplied from an external source into a volume element dV in the deformed solid;

 

· The net heat flux into the solid $Q={\displaystyle \underset{V}{\int }qdV}-{\displaystyle \underset{A}{\int }q\cdot ndA}$

 

· The velocity field in the solid v

 

· The stretch rate $D_{ij}={\displaystyle \frac{1}{2}\left({\displaystyle \frac{\partial v_i}{\partial x_j}+{\displaystyle \frac{\partial v_j}{\partial x_i}}}\right)}$ 

 

· The Cauchy stress distribution in the solid $\sigma$

 

· The net rate of mechanical work done on the solid $W={\displaystyle \underset{V}{\int }\rho b\cdot v}dV+{\displaystyle \underset{A}{\int }n\cdot (\sigma }v)dA$

 

· The total kinetic energy $KE={\displaystyle \underset{V}{\int }{\displaystyle \frac{1}{2}\rho (v_i{v}_i)dV}}$

 

· The total internal energy ${\rm E}={\displaystyle \underset{V}{\int }\rho \epsilon dV}$ where $\epsilon$ is the specific internal energy (internal energy per unit mass)

 

 

· The total entropy $S={\displaystyle \underset{V}{\int }\rho sdV}$, where $s$ is the specific entropy (entropy per unit mass)

 

· The temperature of the solid $\theta$.

 

 

· The net external entropy supplied to the volume ${\displaystyle \frac{d{\rm H}}{dt}={\displaystyle \underset{A}{\int }-{\displaystyle \frac{q}{\theta }\cdot ndA}}+{\displaystyle \underset{V}{\int }{\displaystyle \frac{q}{\theta }dV}}}$

 

· The specific free energy $\psi =\epsilon -\theta s$

 

· The total free energy $\Psi ={\displaystyle \underset{V}{\int }\rho \psi dV}$

 

 

The first law of thermodynamics then requires that

${\displaystyle \frac{d}{dt}({\rm E}+KE)=Q+W}$

for any volume V.  

 

This condition can also be expressed as

$\rho {\left.{\displaystyle \frac{\partial \epsilon }{\partial t}}\right|}_{x=const}={\sigma }_{ij}{D}_{ij}-{\displaystyle \frac{\partial q_i}{\partial y_i}+q}$

 

To see this,

 

1.  Recall that

$W={\displaystyle \underset{V}{\int }\rho b_i{v}_i}dV+{\displaystyle \underset{A}{\int }{\sigma }_{ij}{n}_i{v}_j}dA={\displaystyle \underset{V}{\int }{\sigma }_{ij}{D}_{ij}dV}+{\displaystyle \frac{d}{dt}\left\{{\displaystyle \underset{V}{\int }{\displaystyle \frac{1}{2}\rho v_i{v}_{i}dV}}\right\}={\displaystyle \underset{V}{\int }{\sigma }_{ij}{D}_{ij}dV}+{\displaystyle \frac{d}{dt}(KE)}}$

 

2.  The divergence theorem gives

$Q={\displaystyle \underset{V}{\int }qdV}-{\displaystyle \underset{A}{\int }q\cdot ndA}={\displaystyle \underset{V}{\int }\left(q-{\displaystyle \frac{\partial q_i}{\partial y_i}}\right)dV}$

 

3.  Therefore

${\displaystyle \frac{d}{dt}\left({\displaystyle \underset{V}{\int }\rho \epsilon dV}+KE\right)={\displaystyle \underset{V}{\int }\left(q-{\displaystyle \frac{\partial q_i}{\partial y_i}}\right)dV}+{\displaystyle \underset{V}{\int }{\sigma }_{ij}{D}_{ij}dV}+{\displaystyle \frac{d}{dt}(KE)}}$

 

4.  Note also that

${\displaystyle \frac{d}{dt}{\displaystyle \underset{V}{\int }\rho \epsilon dV}={\displaystyle \frac{d}{dt}{\displaystyle \underset{V_0}{\int }{\rho }_0\epsilon dV}={\displaystyle \underset{V_0}{\int }{\rho }_0{\displaystyle \frac{d\epsilon }{dt}dV}}={\displaystyle \underset{V}{\int }\rho {\displaystyle \frac{d\epsilon }{dt}dV}}}}$

where ${\rho }_0=J\rho$ is the mass density per unit reference volume.

 

5.  Finally

${\displaystyle \underset{V}{\int }\rho {\displaystyle \frac{d\epsilon }{dt}dV}}={\displaystyle \underset{V}{\int }\left(q-{\displaystyle \frac{\partial q_i}{\partial y_i}}\right)dV}+{\displaystyle \underset{V}{\int }{\sigma }_{ij}{D}_{ij}dV}$

 

This must hold for all V, giving the required result.

 

 

The second law of thermodynamics specifies that the net entropy production within V must be non-negative, i.e.

${\displaystyle \frac{dS}{dt}-{\displaystyle \frac{d{\rm H}}{dt}\ge 0}}$

This can also be expressed as

$\rho {\displaystyle \frac{\partial s}{\partial t}+{\displaystyle \frac{\partial (q_i/\theta )}{\partial y_i}-{\displaystyle \frac{q}{\theta }\ge 0}}}$

(this condition is know as the Clausius-Duhem inequality).

 

To see this, simply substitute the definitions of S and $dH/dt$ and use the divergence theorem to re-write the area integral as a volume integral. 

 

 

The first and second laws can be combined to yield the free energy imbalance

${\sigma }_{ij}{D}_{ij}-{\displaystyle \frac{1}{\theta }{q}_i{\displaystyle \frac{\partial \theta }{\partial y_i}-\rho \left({\displaystyle \frac{\partial \psi }{\partial t}+s{\displaystyle \frac{\partial \theta }{\partial t}}}\right)\ge 0}}$

This form of the second law is particularly helpful if you need to check whether a stress-strain law satisfies the second law of thermodynamics.

 

 

To prove the expression for free energy imbalance,

 

1. Start with the second law, and expand the derivative of $q_i/\theta$

$\rho {\displaystyle \frac{\partial s}{\partial t}+{\displaystyle \frac{\partial (q_i/\theta )}{\partial y_i}-{\displaystyle \frac{q}{\theta }=\rho {\displaystyle \frac{\partial s}{\partial t}+{\displaystyle \frac{1}{\theta }{\displaystyle \frac{\partial q_i}{\partial y_i}-{\displaystyle \frac{q}{\theta }-{\displaystyle \frac{1}{{\theta }^2}{q}_i{\displaystyle \frac{\partial \theta }{\partial y_i}}}}}}}}}}$

 

2. Use the first law to substitute for $\partial q_i/\partial y_i$, which shows that

$\begin{array}{l}\rho {\displaystyle \frac{\partial s}{\partial t}+{\displaystyle \frac{1}{\theta }{\displaystyle \frac{\partial q_i}{\partial y_i}-{\displaystyle \frac{q}{\theta }-{\displaystyle \frac{1}{{\theta }^2}{q}_i{\displaystyle \frac{\partial \theta }{\partial y_i}=\rho {\displaystyle \frac{\partial s}{\partial t}+{\displaystyle \frac{1}{\theta }\left(-\rho {\left.{\displaystyle \frac{\partial \epsilon }{\partial t}}\right|}_{x=const}+{\sigma }_{ij}{D}_{ij}\right)-{\displaystyle \frac{1}{{\theta }^2}{q}_i{\displaystyle \frac{\partial \theta }{\partial y_i}\ge 0}}}}}}}}}}\\ \Rightarrow {\sigma }_{ij}{D}_{ij}-\rho {\displaystyle \frac{\partial }{\partial t}(\psi +\theta s)+\theta \rho {\displaystyle \frac{\partial s}{\partial t}-{\displaystyle \frac{1}{\theta }{q}_i{\displaystyle \frac{\partial \theta }{\partial y_i}\ge 0}}}}\end{array}$

 

where we have substituted $\epsilon =\psi +\theta s$ and noted that temperature is always positive.  This yields the solution.

 

 

The free energy imbalance can also be expressed as a condition on the total energy and heat flux into the solid

$W-{\displaystyle \frac{d(KE)}{dt}-{\displaystyle \frac{d\Psi }{dt}-{\displaystyle \underset{V}{\int }\left(\rho s{\displaystyle \frac{\partial \theta }{\partial t}+{\displaystyle \frac{1}{\theta }{q}_i{\displaystyle \frac{\partial \theta }{\partial y_i}}}}\right)dV}\ge 0}}$

This result follows by integrating the local form over the volume V, and using the stress-power work expression (Sect 2.5.1).